Solution to Q#1 For any sets 𝐴, 𝐡 and 𝐢, if 𝐴 βŠ† 𝐡, then 𝐴 ∩ … - Sikademy
Author Image

Archangel Macsika

Q#1 For any sets 𝐴, 𝐡 and 𝐢, if 𝐴 βŠ† 𝐡, then 𝐴 ∩ 𝐢 βŠ† 𝐡 ∩ 𝐢. Q#2 For any sets 𝐴, 𝐡 and 𝐢, (𝐴 Γ— 𝐢) ∩ (𝐡 Γ— 𝐷) = (𝐴 ∩ 𝐡) Γ— (𝐢 ∩ 𝐷). Q#3 Given sets 𝐴, 𝐡 and 𝐢, prove that 𝐴 Γ— (𝐡 ∩ 𝐢) = (𝐴 Γ— 𝐡) ∩ (𝐴 Γ— 𝐢) Q#4 Prove that If 𝐴 and 𝐡 are sets, then 𝒫(𝐴)⋃𝒫(𝐡) βŠ† 𝒫(𝐴⋃𝐡). Q#5 Suppose 𝐴 and 𝐡 are sets. If 𝒫(𝐴) βŠ† 𝒫(𝐡),π‘‘β„Žπ‘’π‘› 𝐴 βŠ† 𝐡.

The Answer to the Question
is below this banner.

Can't find a solution anywhere?


Get the Answers Now!

You will get a detailed answer to your question or assignment in the shortest time possible.

Here's the Solution to this Question


LetΒ x\in A\cap C

Case 1,

SinceΒ A\sube B, x\in A\implies x\in B

x\in B\implies x\in B\cap C

Case 2,

x\in C

x\in C\implies x\in B\cap CΒ sinceΒ x\in B

Thus, in either cases,Β x\in A\cap C\implies x\in B\cap C.


\forall x(x\in A\cap C\implies x\in B\cap C)

We now have that,

A\cap C\sube B\cap CΒ \blacksquare.


LetΒ (x,y)\in(A\cap B)\times(C\cap D)

\iff x\in (A\cap B)\times(C\cap D)\times y

\iff x\in A\cap B\land y\in C\cap D

\iff x\in A\land x\in B\land y\in C\land y\in D

\iff x(A\times C)y\land x(B\times D)y

\iff x(A\times C)\cap(B\times D)y


(A\cap B)\times(C\cap D)=(A\times C )\cap (B\times D)Β as requiredΒ \blacksquareΒ .


To begin with, we will first show thatΒ A\times (B\cap C)\sube (A\times B)\cap(A\times C ). SupposeΒ (x,y)\in A\times(B\cap C). This means that,Β x\in AΒ andΒ y\in B\cap CΒ by definition of Cartesian product. By definition of intersection, it follows thatΒ y\in BΒ andΒ y\in C.SinceΒ x\in AΒ andΒ y\in B, it follows thatΒ (x,y)\in A\times BΒ by definition of Cartesian product. Also, sinceΒ x\in AΒ andΒ y\in CΒ it follows thatΒ (x,y)\in A\times CΒ by definition ofΒ \times. Now, we haveΒ (x,y)\in A\times BΒ andΒ (x,y)\in A\times CΒ and so,Β (x,y)\in (A\times B)\cap(A\times C). Having shown thatΒ (x,y)\in A\times(B\cap C)\implies (x,y)\in (A\times B)\cap(A\times C), we have,Β A\times(B\cap C)\sube(A\times B)\cap(A\times C).

Next, we show thatΒ (A\times B)\cap(A\times C)\sube A \times(B\cap C),

SupposeΒ (x,y)\in (A\times B)\cap(A\times C). By definition of intersection, it means that,Β (x,y)\in A\times BΒ andΒ (x,y)\in A\times C. By definition of Cartesian product,Β (x,y)\in A\times BΒ meansΒ x\in AΒ andΒ y\in BΒ also,Β (x,y)\in A\times CΒ meansΒ x\in AΒ andΒ y\in C. We have that,Β y\in BΒ andΒ y\in CΒ so,Β y\in B\cap CΒ by definition of intersection. Thus, we have deduced thatΒ x\in AΒ andΒ y\in B\cap C, soΒ (x,y)\in A\times (B\cap C). Hence,

(A\times B)\cap(A\times C)\sube A\times (B\cap C).

In summary, we have shown thatΒ A\times (B\cap C)\sube (A\times B)\cap(A\times C)Β andΒ (A\times B)\cap(A\times C)\sube A\times (B\cap C). It follows that,Β A\times(B\cap C)=(A\times B)\cap(A\times C)Β as required\blacksquare.


SupposeΒ x\in \mathcal{p}(A)\cup \mathcal{p}(B),

Then,Β x\in\mathcal{p}(A) \space or\space x\isin\mathcal{p}(B)


x\sube A\space or\space x\sube B

We need to show thatΒ x\in\mathcal{p}( A\cup B),Β that is, we need to show thatΒ x\sube A\cup B

SupposeΒ y\in xΒ then,Β y\in A\space or\space y\in BΒ sinceΒ x\sube A \space or \space x\sube B

Thus,Β y\in x\implies y\in (A\cup B)Β and so,Β x\sube (A\cup B).


x\in\mathcal {p}(A\cup B)

\therefore x\in \mathcal{p}(A)\cup \mathcal{p}(B)\implies x\in\mathcal {p}(A\cup B)

and so,

\mathcal{p}(A)\cup\mathcal{p}(B)\sube \mathcal{p}(A\cup B)Β as requiredΒ \blacksquareΒ .


Suppose thatΒ AΒ andΒ BΒ are sets andΒ \mathcal p(A)\sube\mathcal {p})(B)

By definition of a power set,Β A\sube \mathcal {p}(A).SinceΒ A\in \mathcal{p}(A)Β andΒ \mathcal{p}(A)\sube \mathcal{p}(B), we know thatΒ A\in \mathcal{p}(B)Β by definition of subset.

So, by definition of power set,Β A\sube BΒ \blacksquare.

Related Answers

Was this answer helpful?

Join our Community to stay in the know

Get updates for similar and other helpful Answers

Question ID: mtid-5-stid-8-sqid-1325-qpid-1063